Question

Prove that if n is an integer greater than 2,
then n^3 - 1 is composite.

Answer 1

n^3-1=(n-1){1+n(n+1)}
If n is odd, n-1 is even and n+1 is even. So, n(n+1) is also even.
Thus, n(n+1)+1 is odd. So, multiplication of even no (n-1) and odd n(n+1)+1 gives an even number, which is composite.
If n is even, n-1 is odd and n+1 is odd. So, n(n+1) is even.
Thus, n(n+1)+1 is odd. So, multiplication of odd no (n-1) and odd n(n+1)+1 gives an odd number, which is composite.

Answer 2

good job mani

Answer 3

excellent explanation

Answer 4

Thank you! :)

Answer 5

Thanks guys

Answer 6

If \(n>2\) then \(n-1\geq 2\) which means that

\[n^3-1=(n-1)(n^2+n+1)\]

must be composite.